Question #8e17c

1 Answer
Feb 27, 2017

#ln|(x - 2)/(x - 3)| + ln|x-2|#

Explanation:

Right off the bat, we can see that we will need to use impartial fraction integrals.

We set it as #inta/(x-3)dx + b/(x-2)dx#
#ax + bx = 1#
#a + b = 1#
#-2a - 3b = -4#

Solving it out, we get #a = -1# and #b = 2#

#int(-1)/(x-3)dx+int2/(x-2)dx#
#-int1/(x-3)dx+2int1/(x-2)dx#
Integral this #-int1/(x-3)dx+int1/(x-2)dx+int1/(x-2)dx#

to get this #-ln|x-3| + 2ln|x-2|#

Now to integral #[-ln|x-3| + 2ln|x-2|]_0^1#

#(-ln|1-3| + 2ln|1-2|) - (-ln|0-3| + 2ln|0-2|)#
=#(-ln|-2| + 2ln|-1|) - (-ln|-3| + 2ln|-2|)#
=#(-ln2 + ln1) - (-ln3 + 2ln2)#
=#-ln2 - ln4/3#
=#-ln8/3#
=#-.981#